Question

Consider the matrix \[ M= \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \] Let \(p,q,r,s,a,b,c,d\) be integers such that \[ M^{26}= \begin{bmatrix} p & q \\ r & s \end{bmatrix} \] and \[ \sum_{k=1}^{26} M^k= \begin{bmatrix} a & b \\ c & d \end{bmatrix}. \] Then which of the following statements is (are) TRUE?

• A. There exists a \(2\times2\) invertible matrix \(N\) with real entries such that \[ MN= N \begin{bmatrix} 1 & 1 0 & 1 \end{bmatrix} \]
• B. The value of \(a\) is \(378\)
• C. For any two given integers \(m\) and \(n\), there exist unique integers \(x\) and \(y\) such that \[ px+qy=m \] and \[ rx+sy=n \]
• D. For each positive real number \(t\), the system of linear equations \[ (a+t)x+by=1 \] \[ cx+(d+t)y=-1 \] has a unique solution

Hint:

If: \[ A^2=0 \] then: \[ (I+A)^n=I+nA \] using binomial expansion.

Answer 1

The Correct Option is A

Step 1: Find characteristic polynomial of \(M\).
\[ M= \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \] Characteristic polynomial: \[ |M-\lambda I| = \begin{vmatrix} 2-\lambda & -1 \\ 1 & -\lambda \end{vmatrix} \] \[ =(2-\lambda)(-\lambda)+1 \] \[ =\lambda^2-2\lambda+1 \] \[ =(\lambda-1)^2 \] Thus eigenvalue: \[ \lambda=1 \] with algebraic multiplicity \(2\).

Step 2: Check Option (A).
Since \(M\) has repeated eigenvalue \(1\) and \[ M\ne I, \] its Jordan form is: \[ \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \] Hence there exists an invertible matrix \(N\) such that: \[ MN= N \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \] Therefore: \[ \Rightarrow \mathrm{Option\ (A)\ is\ Correct} \]

Step 3: Find formula for \(M^n\).
Write: \[ M=I+ \begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix} \] Let: \[ A= \begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix} \] Then: \[ A^2=0 \] Hence: \[ M^n=(I+A)^n=I+nA \] Thus: \[ M^n= \begin{bmatrix} 1+n & -n \\ n & 1-n \end{bmatrix} \] For: \[ n=26 \] \[ M^{26}= \begin{bmatrix} 27 & -26 \\ 26 & -25 \end{bmatrix} \] Thus: \[ p=27,\quad q=-26,\quad r=26,\quad s=-25 \]

Step 4: Check Option (B).
\[ \sum_{k=1}^{26}M^k = \sum_{k=1}^{26} \begin{bmatrix} 1+k & -k \\ k & 1-k \end{bmatrix} \] Now: \[ a=\sum_{k=1}^{26}(1+k) \] \[ =26+\frac{26\cdot27}{2} \] \[ =26+351 \] \[ =377 \] Thus: \[ a\ne378 \] Therefore: \[ \Rightarrow \mathrm{Option\ (B)\ is\ Incorrect} \]

Step 5: Check Option (C).
Determinant of \[ M^{26} \] is: \[ \det(M^{26})=(\det M)^{26} \] Now: \[ \det M=1 \] Hence: \[ \det(M^{26})=1 \] Thus the matrix is invertible over integers. Therefore for every integers \(m,n\), unique integers \(x,y\) exist. Hence: \[ \Rightarrow \mathrm{Option\ (C)\ is\ Correct} \]

Step 6: Check Option (D).
Coefficient matrix: \[ \begin{bmatrix} a+t & b \\ c & d+t \end{bmatrix} \] Its determinant: \[ =(a+t)(d+t)-bc \] Using: \[ \sum_{k=1}^{26}M^k = \begin{bmatrix} 377 & -351 \\ 351 & -325 \end{bmatrix} \] Determinant: \[ =(377+t)(-325+t)+351^2 \] \[ =t^2+52t+2 \] For: \[ t>0, \] this is always positive. Hence determinant never vanishes. Therefore unique solution always exists. Thus: \[ \Rightarrow \mathrm{Option\ (D)\ is\ Correct} \]

Step 7: Identify the correct options.
Hence: \[ \boxed{\mathrm{(A),\ (C)\ and\ (D)}} \]